If the equation x4+ax3−13x2+bx−4=0 (a,b∈R) has one repeated root and one more root being 2+5 , then

The given equation is f(x)=x4+ax3−13x2+bx−4=0Given that one of the roots is 2+5roots are 2+5, 2−5,α,α⇒(2+5)(2−5)α.α=−4 α=±2−−−(1) ∑αβ=−13→α2+8α+12=0→α=−2,α=−6−−(2)from (1) & (2)→α=−2 , a=2+5+2−5−2−2⇒a=0 f(−2)=0⇒b=−20

If the equation x4+ax3−13x2+bx−4=0 (a,b∈R) has one repeated root and one more root being 2+5 , then

The given equation is f(x)=x4+ax3−13x2+bx−4=0Given that one of the roots is 2+5roots are 2+5, 2−5,α,α⇒(2+5)(2−5)α.α=−4 α=±2−−−(1) ∑αβ=−13→α2+8α+12=0→α=−2,α=−6−−(2)from (1) & (2)→α=−2 , a=2+5+2−5−2−2⇒a=0 f(−2)=0⇒b=−20

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If the equation $x^{4} + a x^{3} - 13 x^{2} + b x - 4 = 0$ $(a, b \in R)$ has one repeated root and one more root being $2 + \sqrt{5}$ , then

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$a = 0, b = 10$

$a = 0, b = - 20$

repeated root is 3

repeated root is 2

answer is B.

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Detailed Solution

The given equation is $f (x) = x^{4} + a x^{3} - 13 x^{2} + b x - 4 = 0$

Given that one of the roots is $2 + \sqrt{5}$
roots are $2 + \sqrt{5}, 2 - \sqrt{5}, α, α \Rightarrow (2 + \sqrt{5}) (2 - \sqrt{5}) α . α = - 4$
$α = \pm 2 - - - (1)$
$\sum α β = - 13 \to α^{2} + 8 α + 12 = 0 \to α = - 2, α = - 6 - - (2)$
from $(1) & (2) \to α = - 2, a = 2 + \sqrt{5} + 2 - \sqrt{5} - 2 - 2 \Rightarrow a = 0$

$f (- 2) = 0 \Rightarrow b = - 20$